# Voltage Drop Over Voltmeter

1. Feb 29, 2012

1. The problem statement, all variables and given/known data
Voltmeter has a internal resistance of 10M ohms and the resistor has a resistance of 100,000ohm. The current through the resistor is 1mA.
a)What is the voltage drop across the voltmeter?
b)What is the current through the voltmeter?

2. Relevant equations
V=IR is all you should need

3. The attempt at a solution
I am missing something really small (have taken the class already) but I'm stuck. Im stuck with how to apply the 1mA current. Do I need to use that to find the voltage of the source? Or do I find the overall resistance of the reistor and parallel voltmeter then use the 1mA somehow?

2. Feb 29, 2012

### BruceW

You don't need to find the voltage of the source, or the combined resistance of the parallel resistor and voltmeter. Its simpler than that. You are given the resistance and current through the resistor, so what can you say about the voltage drop over the resistor? And what does this tell you about the voltage drop over the voltmeter?

3. Feb 29, 2012

Well given 1ma over the resistor you can do V = (1E-3)(100,000) which is the voltage drop over the resistor...without the total voltage how can I find the voltage drop over the volt meter? current through voltmeter?

4. Feb 29, 2012

### consciousness

Hint: Potential Difference across all parallel resistors is same.

5. Feb 29, 2012

so if its V=(1E-3)(100,000) = 100 v across the resistor its 100 V across the voltmeter for part a?

For b you then take 100v=I(10E7) = 1E-6 A?

6. Feb 29, 2012

### consciousness

Yes. Note that the voltage refers to potential difference and not always to source voltage

7. Feb 29, 2012

### BruceW

You got part (a) right. And I think you've got the right idea for part (b), but I think you went wrong in part (b). Check you have the right power which 10 is raised to. And write out the equations properly! It makes them easier for you and us to read when you write them out properly. (And your teacher will be able to read them easier, more importantly).

8. Feb 29, 2012